CHAPTER 4Analysis of Variance

One-way ANOVA

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IntroductionIntroduction In this chapter, expand the idea of hypothesis

tests. We describe a test of variances and then a test that simultaneously compares several means to determine if they came from equal populations.

The simultaneous comparison of several population means called analysis of variance (ANOVA).

Using F-test where test whether two samples are populations having equal variances and compare several population means simultaneously.

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ANOVA assumptions:- The population follows the normal

distribution.- The populations have equal

standard deviation.- The populations are independent.

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One-way ANOVA

The one-way analysis of variance specifically allows us to compare several groups of observations whether or not their population mean are equal. One way ANOVA is also known as Completely Randomized Design (CRD).

This design only involves one factor.The application of one way ANOVA requires that the

following assumptions hold true:(i) The populations from which the samples are drawn are

(approximately) normally distributed. (ii) The populations from which the samples are

drawn have the same variance. (iii) The samples drawn from different populations

are random and independent.4

The is the total of all observations from the treatment, while

is the grand total of all N observations.

i .Ythi

Y ..

Treatment

1 2 … i … k

. . . .

. . . .

. . . .

Total

11y

12y

11ny

1Y

21y

22y

22ny

2Y

1iy

2iy

iiny

iY

1ky

2ky

kkny

kY Y

5

Then the hypothesis can be written as:For model :

For model :

6

0 1 2

1

: all the population mean are equal

: for at least one at least one of the mean is not equal

k

i j

H ...

H i, j

ij i ijy

ij i i ijy ,

0 1 2

1

: 0 there is no treatment effect

: 0 for at least one there is exist treatment effect

k

i

H ...

H i

The computations for an analysis of variance problem are usually

summarized in tabular form as shown in table below. This table is

referred to as the ANOVA table.Source of Variation

Sum of Squares

Degree of

freedom

Mean Square F Calculate

d

Treatment

(Between levels)

SSTR k - 1

Error(within levels)

SSE N - k

Total SST N - 1

1

SSTRMSTR

k

cal

MSTRF

MSE

SSEMSE

N k

7

where

We reject if and conclude that some of the data

is due to differences in the treatment levels.

2 222

1 1 1

number of treatment

total number of observation

k n ki. ..

iji j i i

Y YY ..SST y SSTR SSE SST SSTR

N n N

k

N

0H 1cal ,k ,N kF f

1,k ,N kf

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Example 4.2Three different types of acid can be used in a particular

chemical process.The resulting yield (in %) from several batches using the

different types of acid are given below:

Test whether or not the three populations appear to have equal means using = 0.05.

Acid

A B C

93 95 76

95 97 77

74 87 84

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Solution:

1. Construct the table of calculation:

2. Set up the hypothesis:

9 3N ,k

Acid

A B C

93 95 76

95 97 77

74 87 84

1 262.Y 2 279.Y 3 237.Y 778..Y

0

1

:

: for at least one A B C

i j

H

H i, j

10

3. Construct ANOVA table:

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1 1

2

2 2 2 2 2 2

2 2

1

2 2 2 2

778 93 95 74 76 77 84

9 67914 67253 7778

660 2222

262 279 237 778

3 3 3 9

67551 3333 67253 7778

k n..

iji j

ki . ..

i i

YSST y

N

...

.

.

Y YSSTR

n N

. .

297 5555

660 2222 297 5555

362 6667

.

SSE SST SSTR

. .

.

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4. At = 0.05, from the statistical table for f distribution, we have

5. Since , thus we failed to reject and conclude that there is no difference for mean in the three types of acid at significance at = 0.05

Source of Variation

Sum of Squares

Degrees of

Freedom

Mean Square F Calculated

Treatment 297.5555

3 – 1 = 2

Error 362.6667

9 – 3 = 6

Total 660.2222

9 – 1 = 8

297 5555148 7778

2

..

362 666760 4445

6

..

148 77782 4614

60 4445cal

.F .

.

0 05 2 6 5 14. , ,f .

0 05 2 62 4614 5 14cal . , ,F . f . 0H

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Exercise 4.1:Four catalyst that may affect the concentration of one component in

a three-component liquid mixture are being investigated. The following concentrations are obtained.

Compute a one-way analysis of variance for this experiment and test the hypothesis at 0.05 level of significance and state your conclusion concerning the effect of catalyst on the concentration of

one component in three-component liquid mixture.

1 2 3 4

58.2 56.3 50.1 52.9

57.2 54.5 54.2 49.9

58.4 57.0 55.4 50.0

55.8 55.3 51.7

54.9

1Y 2Y 3Y 4Y Y

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Exercise 4.2:

The following is a partial ANOVA table:

Complete the table and answer the following questions:

a)Complete the ANOVA table. (red colour is the answer)

b)Write all the hypothesis

c)What is your conclusion regarding the hypothesis? Using 5%significance level.

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Source Sum of Square

s

df Mean Square

F

Treatment

500-180 = 320

2 320 / 2 = 160

160 / 20 = 8

Error 9 * 20 =180

11-2=9 20

Total 500 11

Exercise 4.3

The following is sample information. Test the hypothesis that the treatment means are equal. Use . (Answer: Reject )

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0.05

Treatment 1 Treatment 2 Treatment 3

8 3 3

6 2 4

10 4 5

9 3 4

0H